If $f'(x)=2f(x)$ and $f(1)=5$, then $f(3)=me^n$ for some integers $m$ and $n$. What are $m$ and $n$ ? $m=~$
Solution: If we let $y=f(x)$ and change the derivative notation, we can rewrite the differential equation so that it's separable. $\dfrac{dy}{dx}=2y$ What does it look like after we separate the variables? $\dfrac1y\,dy=2\,dx$ Let's integrate both sides of the equation. $\int\dfrac1y\,dy=\int2\,dx$ What do we get? $\ln |y|=2x+C$ What value of $C$ satisfies the initial condition $y(1)=5$ ? Let's substitute $x=1$ and $y=5$ into the equation and solve for $C$. $\begin{aligned} \ln |5| &= 2\cdot1+C\\ \\ C&=-2+\ln 5 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \ln |y|&=2x-2+\ln 5\\ \\ |y|&=e^{2x-2+\ln 5}\\ \\ |y|&=e^{2x-2}\cdot e^{\ln 5}\\ \\ |y|&=5e^{2x-2}\\ \\ y&=\pm5e^{2x-2} \end{aligned}$ Which sign satisfies the initial condition $y(1)=5$ ? We must choose the positive sign to satisfy the initial condition. $y=5e^{2x-2}$ If $y(3)=me^n$, what are the integers $m$ and $n$ ? $\begin{aligned} y(3)&=5e^{2\cdot3-2}\\ &=5e^4\\ &=me^n \end{aligned}$ Thus, $m=5$ and $n=4$.